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I am trying to convert the following problem to polar form:

$$z=-j10.$$

Using this equation, where $|z|=r=\sqrt{x^2+y^2}$ and $\arg z=\theta=\arctan(y/x).$

$$\eqalign{z&=|z|e^{j\arg z}\\ &=re^{j\theta}\\&=r\angle\theta.}$$

I determined that x = 0 and y = -10. However, if I plug x and y into arctan(y/x), the result would be indetermined since we're dividing by 0. The solution to that problem is 10<-90degrees.

Could someone give me some insight on how to convert the above cartesian to polar form?

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    $\begingroup$ Above all and before anything else, you should draw the picture. $\endgroup$ – Lubin Jun 30 '14 at 19:09
  • $\begingroup$ If you draw the Argand diagram for this, the answer should be immediately apparent. $\endgroup$ – Myridium Jun 30 '14 at 19:19
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From

http://hotmath.com/hotmath_help/topics/polar-form-of-a-complex-number.html

The polar of a complex number is given by: $$z = r(\cos(\theta) + i\sin(\theta))$$ In your example: $$z = -10i$$ $$r = \sqrt{0^2+(-10)^2} = 10$$ $$\theta = \arctan(\frac{-10}{0}) = \frac{3\pi}{2}$$ $\theta$ is $\frac{3\pi}{2}$ because the complex number is in the III quadrant. So the polar form of our complex number is $$z = 10(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}))$$

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  • $\begingroup$ Thanks Vishwa! This answer really explained it in the way I understand. $\endgroup$ – theGreenCabbage Jun 30 '14 at 19:41
  • $\begingroup$ you're welcome! $\endgroup$ – Vishwa Iyer Jun 30 '14 at 19:42
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First find $\alpha$ that $\cos \alpha=0$, there are two values in $[0,2\pi)$:$\frac{\pi}{2}$ and $\frac{3\pi}{2}$, for second you have $\sin \alpha=-1$ for first $\sin \alpha=1$, so: $-10i=10(\cos \frac{3\pi}{2} + i\sin\frac{3\pi}{2})$

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  • $\begingroup$ Sorry, but how did you get the 3pi/2? $\endgroup$ – theGreenCabbage Jun 30 '14 at 19:10
  • $\begingroup$ Hi agha. Thank you for your assistance. Would I simply use Euler's Formula to convert your form to the polar form I want? $\endgroup$ – theGreenCabbage Jun 30 '14 at 19:16
  • $\begingroup$ Yes, but in this case it's easier, because $x=0$, so you can get solution immediately. $\endgroup$ – agha Jun 30 '14 at 19:18

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